Distance from Alexander City to Montechiarugolo
The shortest distance (air line) between Alexander City and Montechiarugolo is 4,944.89 mi (7,958.03 km).
How far is Alexander City from Montechiarugolo
Alexander City is located in Alabama, United States within 32° 55' 22.44" N -86° 3' 51.12" W (32.9229, -85.9358) coordinates. The local time in Alexander City is 01:04 (15.02.2025)
Montechiarugolo is located in Parma, Italy within 44° 41' 36.24" N 10° 25' 20.64" E (44.6934, 10.4224) coordinates. The local time in Montechiarugolo is 08:04 (15.02.2025)
The calculated flying distance from Montechiarugolo to Montechiarugolo is 4,944.89 miles which is equal to 7,958.03 km.
Alexander City, Alabama, United States